Proof

So it remains to check that for an essentially surjective ff, being 0-connected is equivalent to being full.

The homotopy pullbackX×YXX \times_Y X is given by the groupoid whose objects are triples (x1,x2∈X,α:f(x1)→f(x2))(x_1, x_2 \in X, \alpha : f(x_1) \to f(x_2)) and whose morphisms are corresponding tuples of morphisms in XX making the evident square in YY commute.

By prop. it is sufficient to check that the diagonal functor X→X×YXX \to X \times_Y X is (-1)-connected, hence, as before, essentially surjective, precisely if ff is full.

First assume that ff is full. Then for (x1,x2,α)∈X×YX(x_1,x_2, \alpha) \in X \times_Y X any object, by fullness of ff there is a morphism α^:x1→x2\hat \alpha : x_1 \to x_2 in XX, such that f(α^)=αf(\hat \alpha) = \alpha.

Conversely, assume that the diagonal is essentially surjective. Then for every pair of objects x1,x2∈Xx_1, x_2 \in X such that there is a morphism α:f(x1)→f(x2)\alpha : f(x_1) \to f(x_2) we are guaranteed morphisms h1:x1→x2h_1 : x_1 \to x_2 and h2:x2→x2h_2 : x_2 \to x_2 such that